已知tan(-a-4π/3)=-5,则tan(π/3+a)的值为

问题描述:

已知tan(-a-4π/3)=-5,则tan(π/3+a)的值为

因为tan(-a)=-tana,
tan(π+a)=tana
已知tan(-a-4π/3)=-tan(-a-4π/3)=-5,
则tan(4π/3+a)=5
tan(4π/3+a)=5
tan(π+π/+a)=5
所以,tan(π/3+a)=5


tan(-a-4π/3)
=tan[-(a+4π/3)]
=-tan(a+4π/3)
=-tan(a+π+π/3)
=-tan(a+π/3)
=-5

∴tan(a+π/3)=5